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3 years ago

Consider the following reaction:

Chemistry

1 answer:

vfiekz [6]3 years ago

3 0

Answer:

Kc = 4.76

Explanation:

To find the concentrations of CO and H₂ at equilibrium, you have to set up an ICE (Initial, Change, Equilibrium) table.

CO (g) + 2H₂(g) ⇌ CH₃OH (g)

I 0.32 M 0.53 M 0

C -x -2x +x

E 0.32-x 0.53-2x 0.16 M

Since you know the concentration of CH₃OH at equilibrium, it would be equal to x since 0 + x = 0.16. So,

[CH₃OH] = 0.16 M

[CO] = 0.32 - 0. 16 = 0.16 M

[H₂] = 0.53 - 2(0.16) = 0.21 M

Now that you have all the concentrations at equilibrium, you can calculate the equilibrium constant.

Kc = products ÷ reactants

= [CH₃OH] ÷ [CO][H₂]²

= 0.16 ÷ (0.16)(0.21)

Kc = 4.76

The equilibrium constant at this temperature is 4.76.

Hope this helps.

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The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per uni

hjlf

Answer:

Turnover number = 113182s⁻¹

Explanation:

Turnover number is a concept used to know the activity of an enzyme. The higher turnover number, the higher activity of the enzime.

Turnover number is defined as:

Turnover number = Rmax / [E]t

As Rmax is 249 μmol⋅L⁻¹⋅s⁻¹ and [E]t is 2.20 nmol⋅L⁻¹

As 1μmol = 1000nmol, Rmax = 249000 nmol⋅L⁻¹⋅s⁻¹. Replacing:

Turnover number = 249000 nmol⋅L⁻¹⋅s⁻¹ / 2.20 nmol⋅L⁻¹

<h3>Turnover number = 113182s⁻¹</h3>

8 0

3 years ago

How many grams of NO are required to produce 145 g of N2 in the following reaction?

V125BC [204]

Answer:

b. 186 g

Explanation:

Step 1: Write the balanced equation.

4 NH₃(g) + 6 NO(g) → 5 N₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 145 g of N₂

The molar mass of nitrogen is 28.01 g/mol.

$145g \times \frac{1mol}{28.01 g} =5.18 mol$

Step 3: Calculate the moles of NO required to produce 5.18 moles of N₂

The molar ratio of NO to N₂ is 6:5.

$5.18molN_2 \times \frac{6molNO}{5molN_2} = 6.22molNO$

Step 4: Calculate the mass corresponding to 6.22 moles of NO

The molar mass of NO is 30.01 g/mol.

$6.22mol \times \frac{30.01g}{mol} =186 g$

4 0

3 years ago

Onur daima düz ve kendisinden küçük görüntü elde etmek istiyor

KIM [24]

It would probably be A

5 0

3 years ago

A student determines the aluminum content of a solution by first precipitating it as aluminum hydroxide, and then decomposing th

Vladimir [108]

Answer: The student should obtain <u>1.103 g of aluminum oxide </u>

Explanation:

First we write down the equations that represent the aluminum hydroxide precipitation from the reaction between the aluminum nitrate and the sodium hydroxide:

Al(NO3)3 + 3NaOH → 3NaNO3 + Al(OH)3

Now, the equation that represents the decomposition of the hydroxide to aluminum oxide by heating it.

2Al(OH)3 → Al2O3 + 3H2O

Second, we gather the information what we are going to use in our calculations.

Volumen of Al(NO3)3 = 40mL

Molar concentration of Al(NO3)3 = 0.541M

Molecular Weight Al2O3 = 101.96 g/mol

Third, we start using the molar concentration of the aluminum nitrate and volume used to find out the total amount of moles that are reacting

$\frac{0.541moles Al(NO3)3}{1L} x\frac{1L}{1000mL} x 40mL Al(NO3)3 = 0.022moles Al(NO3)3$

then we use the molar coefficients from the equations to discover the amount of Al2O3 moles produced

$0.022moles Al(NO3)3 x \frac{1mol Al(OH)3}{1mol Al(NO3)3} X\frac{1mol Al2O3}{2molAl(OH)3} = 0.011 moles Al2O3$

finally, we use the molecular weight of the Al2O3 to calculate the final mass produced.

$0.011moles Al2O3 x \frac{101.96g Al2O3}{1mol Al2O3} = 1.103g Al2O3$

4 0

4 years ago

Write about buffer solution with its formulas

musickatia [10]

Answer:

A buffer solution is a mixture of two pair salts to maintain a stable pH.

Explanation:

This mixture is made always with a complementary pair of salts (one acid an another basic).

Example:

Acetic Acid (CH3COOH) and Sodium Acetate (NaH3COO).

So when you add a little bit more acid to this mixture, the basic part of the buffer, in this case the Sodium Acetate, will neutralize it and the pH will remain the same.

On the other hand, if the mixture receives some basic substance, the acid part of the buffer, the acetic acid, will neutralize it, so again the pH will remain the same.

8 0

4 years ago